Question

If ${\cos }^{4}x-(\lambda +2){\cos }^{2}x-(\lambda +3)=0$ has a solution, then

• A. $\lambda \ge -3$
• B. $-2\le \lambda \le 0$
• C. $-3\le \lambda \le -2$
• D. $-3\le \lambda \le 0$

Answer 1

The correct option is C $-3\le \lambda \le -2$

${\cos }^{4}x-(\lambda +2){\cos }^{2}x-(\lambda +3)=0\Rightarrow {\cos }^{2}x=\frac{(\lambda +2)\pm \sqrt{(\lambda +2{)}^2+4(\lambda +3)}}{2}\Rightarrow {\cos }^{2}x=\frac{(\lambda +2)\pm \sqrt{{\lambda }^2+4\lambda +4+4\lambda +12}}{2}\Rightarrow {\cos }^{2}x=\frac{(\lambda +2)\pm \sqrt{(\lambda +4{)}^2}}{2}\Rightarrow {\cos }^{2}x=\lambda +3,-1$
As ${\cos }^{2}x\in [0,1]$, so
${\cos }^{2}x=\lambda +3\Rightarrow \lambda +3\in [0,1]\therefore \lambda \in [-3,-2]$